Это решение дает правильный результат. Пока имеются индексы, это должно работать нормально для вашего набора производственных данных, но я проверял его не больше, чем на ваших выборочных данных.
DECLARE @tree TABLE
(ID INT
,name VARCHAR(15)
,ParentID INT
,TYPE TINYINT
)
DECLARE @salary TABLE
(ID INT
,Salary INT
)
INSERT @tree
SELECT 1,'PA',NULL,0
UNION SELECT 2,'Pittsburgh',1,1
UNION SELECT 3,'Accounts',2,1
UNION SELECT 4,'Alex',3,2
UNION SELECT 5,'Robin',3,2
UNION SELECT 6,'HR',2,1
UNION SELECT 7,'Robert',6,2
INSERT @salary
SELECT 4,6000
UNION SELECT 5,5000
UNION SELECT 7,4000
;WITH salaryCTE
AS
(
SELECT t.*
,s.Salary
FROM @tree AS t
LEFT JOIN @salary AS s
ON s.ID = t.ID
)
,recCTE
AS
(
SELECT t.ID
,CAST(t.name AS VARCHAR(MAX)) AS name
,t.ParentID
,ISNULL(t.Salary,0) AS Salary
,0 AS LEVEL
,CAST(t.ID AS VARCHAR(100)) AS ord
FROM salaryCTE AS t
WHERE t.ParentID IS NULL
UNION ALL
SELECT t.ID
,CAST(REPLICATE(' ',r.LEVEL) + t.name AS VARCHAR(MAX)) AS name
,t.ParentID
,ISNULL(t.Salary,0) AS Salary
,r.LEVEL + 1
,CAST(r.ord + '|' + CAST(t.ID AS VARCHAR(11)) AS VARCHAR(100)) AS ord
FROM salaryCTE AS t
JOIN recCTE AS r
ON r.ID = t.ParentID
)
SELECT name
,salary
FROM (
SELECT r1.name
,r1.ord
,SUM(r2.salary) AS salary
FROM recCTE AS r1
LEFT JOIN recCTE AS r2
ON r2.ord LIKE r1.ord + '%'
GROUP BY r1.name,r1.ord
) AS x
ORDER BY ord,name